Question

A solid ball of radius R has a charge density \(\rho\) given by \(\rho=\rho_0\left(1-\frac{r}{R}\right)\) for \(0\le r\le R\). The electric field outside the ball is :

• A. \(\frac{\rho_0R^3}{\varepsilon_0r^2}\)
• B. \(\frac{\rho_0R^3}{12\,\varepsilon_0r^2}\)
• C. \(\frac{4\rho_0R^3}{3\,\varepsilon_0r^2}\)
• D. \(\frac{3\rho_0R^3}{4\,\varepsilon_0r^2}\)

Hint:

Integrate \(\rho\,4\pi r^2dr\) from 0 to R for total charge, then use Gauss law outside.

Answer 1

The Correct Option is B

Total charge: \(Q=\int_0^R\rho_0\left(1-\frac{r}{R}\right)4\pi r^2\,dr=4\pi\rho_0\left[\frac{R^3}{3}-\frac{R^3}{4}\right]=\frac{\pi\rho_0R^3}{3}\).
Outside the ball (Gauss law): \(E\cdot4\pi r^2=\frac{Q}{\varepsilon_0}\).
\(E=\frac{Q}{4\pi\varepsilon_0r^2}=\frac{\rho_0R^3}{12\,\varepsilon_0r^2}\).
Answer: (2) \(\frac{\rho_0R^3}{12\,\varepsilon_0r^2}\)